9 Images Formed by a Converging Lens Interactive

Example 36.9 Images Formed by a Converging Lens Interactive

A converging lens of focal length 10.0 cm forms images of

Solution

objects placed (A) First we construct a ray diagram as shown in Figure

(A)

30.0 cm, 36.30a. The diagram shows that we should expect a real,

inverted, smaller image to be formed on the back side of the

(B)

10.0 cm, and lens. The thin lens equation, Equation 36.16, can be used to

(C)

5.00 cm from the lens.

find the image distance:

In each case, construct a ray diagram, find the image

distance and describe the image.

SECTION 36.4• Thin Lenses

1 I, F

F 1 5.00 cm

Figure 36.30 (Example 36.9) An image is formed by a converging lens. (a) The object is farther from the lens than the focal point. (b) The object is closer to the lens than the focal point.

1 1 1 and the magnification of the image is &

The negative image distance tells us that the image is virtual The positive sign for the image distance tells us that the

and formed on the side of the lens from which the light is image is indeed real and on the back side of the lens. The

incident, the front side. The image is enlarged, and the posi- magnification of the image is

tive sign for M tells us that the image is upright. q

15.0 cm

M"$ "$

What If? What if the object moves right up to the lens p

surface, so that p B 0? Where is the image? Thus, the image is reduced in height by one half, and the

30.0 cm

Answer In this case, because p ** R , where R is either of the negative sign for M tells us that the image is inverted.

radii of the surfaces of the lens, the curvature of the lens can (B) No calculation is necessary for this case because

be ignored and it should appear to have the same effect as a we know that, when the object is placed at the focal

plane piece of material. This would suggest that the image is point, the image is formed at infinity. This is readily

just on the front side of the lens, at q " 0. We can verify this verified by substituting p " 10.0 cm into the thin lens

mathematically by rearranging the thin lens equation: equation.

(C) We now move inside the focal point. The ray diagram

in Figure 36.30b shows that in this case the lens acts as a magnifying glass; that is, the image is magnified, upright,

If we let p : 0, the second term on the right becomes very large compared to the first and we can neglect 1/f. The

on the same side of the lens as the object, and virtual.

equation becomes

Because the object distance is 5.00 cm, the thin lens equa- tion gives

q"$p"0

10.0 cm

Thus, q is on the front side of the lens (because it has the

opposite sign as p), and just at the lens surface. Investigate the image formed for various object positions and lens focal lengths at the Interactive Worked Example link at

q"

http://www.pse6.com.

CHAPTER 36• Image Formation